Suppose `a , b , c` are such that the curve `y=a x^2+b x+c` is tangent to `y=3x-3a t(1,0)` and is also tangent to `y=x+1a t(3,4)dot` Then the value of `(2a-b-4c)` equals ______

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) such that the curve \( y = ax^2 + bx + c \) is tangent to the lines \( y = 3x - 3 \) at the point \( (1, 0) \) and \( y = x + 1 \) at the point \( (3, 4) \). Finally, we will calculate \( 2a - b - 4c \).

### Step 1: Set up the equations from the tangency conditions

1. **For the first tangent line \( y = 3x - 3 \) at the point \( (1, 0) \)**:
- The curve must pass through this point:
\[ 0 = a(1)^2 + b(1) + c \implies a + b + c = 0 \quad \text{(Equation 1)} \]
- The slope of the curve at this point must equal the slope of the line:
\[ \frac{dy}{dx} = 2a(1) + b = 3 \implies 2a + b = 3 \quad \text{(Equation 2)} \]

2. **For the second tangent line \( y = x + 1 \) at the point \( (3, 4) \)**:
- The curve must pass through this point:
\[ 4 = a(3)^2 + b(3) + c \implies 9a + 3b + c = 4 \quad \text{(Equation 3)} \]
- The slope of the curve at this point must equal the slope of the line:
\[ \frac{dy}{dx} = 2a(3) + b = 1 \implies 6a + b = 1 \quad \text{(Equation 4)} \]

### Step 2: Solve the system of equations

Now we have a system of four equations:

1. \( a + b + c = 0 \) (Equation 1)
2. \( 2a + b = 3 \) (Equation 2)
3. \( 9a + 3b + c = 4 \) (Equation 3)
4. \( 6a + b = 1 \) (Equation 4)

#### Step 2.1: Solve Equations 2 and 4 for \( b \)

From Equation 2:
\[ b = 3 - 2a \quad \text{(Substituting into Equation 4)} \]

Substituting \( b \) into Equation 4:
\[ 6a + (3 - 2a) = 1 \implies 4a + 3 = 1 \implies 4a = -2 \implies a = -\frac{1}{2} \]

#### Step 2.2: Find \( b \)

Substituting \( a \) back into Equation 2:
\[ b = 3 - 2(-\frac{1}{2}) = 3 + 1 = 4 \]

#### Step 2.3: Find \( c \)

Substituting \( a \) and \( b \) into Equation 1:
\[ -\frac{1}{2} + 4 + c = 0 \implies c = -\frac{1}{2} - 4 = -\frac{9}{2} \]

### Step 3: Calculate \( 2a - b - 4c \)

Now that we have \( a = -\frac{1}{2} \), \( b = 4 \), and \( c = -\frac{9}{2} \):

\[ 2a - b - 4c = 2(-\frac{1}{2}) - 4 - 4(-\frac{9}{2}) \]
\[ = -1 - 4 + 18 = 13 \]

### Final Answer

Thus, the value of \( 2a - b - 4c \) is \( \boxed{13} \).